Hello!
I have problem:
Derive the solution of the ordinary differential equationd2y/dx2 =f(x), x>0, y(x)=0, dy/dx (0)=0,
in form y(x)= integral from 0 to x [(x-t)f(t)dt].
tnxs

Aug 13th 2007, 10:48 AM

CaptainBlack

Quote:

Originally Posted by perfect

Hello!
I have problem:
Derive the solution of the ordinary differential equationd2y/dx2 =f(x), x>0, y(x)=0, dy/dx (0)=0,
in form y(x)= integral from 0 to x [(x-t)f(t)dt].
tnxs

Some clarification is needed. It is not possible in general to write the solution of:

with , in the form

.

To show this just put .

RonL

Aug 17th 2007, 12:48 AM

Rebesques

perfect:

Quote:

Derive the solution of the ordinary differential equation
d2y/dx2 =f(x), x>0, y(x)=0, dy/dx (0)=0,
in form y(x)= integral from 0 to x [(x-t)f(t)dt]

Are you asked to actually prove the formula, or just show this is the solution? Because the latter is easy: Differentiate twice (under the integral sign) to get y"=f, and since y also gives y(0)=0, y'(0)=0, from the uniqueness theorem this is the only solution!

Yes, I know it's cheating, but it's a way. Now if you are asked to derive the formula, write

and integrating by parts,

, (1)

since y''=f. Now one integration of the differential equation gives us , and substitute into (1) to get

.

Captainblack:

Quote:

To show this just put...

The function must also satisfy the initial conditions.

Aug 17th 2007, 11:49 AM

CaptainBlack

Quote:

Captainblack:

Quote:

To show this just put...

The function must also satisfy the initial conditions.

In fact I was misreading the question statement in another way (the function does not have to satisfy the initial condition but the solution does).